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JEE Physics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

434 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 191–200 of 434
Topics in JEE Physics
Q.191 Medium Optics
A concave mirror has a focal length of 20 cm. An object is placed at the center of curvature. Where will the image form?
A At the center of curvature
B At the focus
C At infinity
D Between focus and pole
Correct Answer:  A. At the center of curvature
EXPLANATION

When object is at center of curvature (u = R = 2f), image forms at the same position (v = R). Magnification = -1

Test
Q.192 Medium Optics
Two slits separated by 0.5 mm are illuminated by light of wavelength 500 nm. The distance to the screen is 1 m. Find the fringe width.
A 0.5 mm
B 1 mm
C 2 mm
D 0.25 mm
Correct Answer:  B. 1 mm
EXPLANATION

Fringe width β = λD/d = (500×10⁻⁹ × 1)/(0.5×10⁻³) = 1×10⁻³ m = 1 mm

Test
Q.193 Medium Magnetism
The SI unit of magnetic flux density (magnetic field) is:
A Weber (Wb)
B Tesla (T)
C Gauss (G)
D Henry (H)
Correct Answer:  B. Tesla (T)
EXPLANATION

The SI unit of magnetic field (flux density) is Tesla (T). 1 T = 1 Wb/m² = 1 kg/(A·s²). Weber is the unit of magnetic flux, Gauss is CGS unit, and Henry is unit of inductance.

Test
Q.194 Medium Magnetism
A long straight wire carrying current I produces a magnetic field at distance r. If the current is doubled and distance is halved, the magnetic field becomes:
A 2 times
B 4 times
C 8 times
D Same as before
Correct Answer:  B. 4 times
EXPLANATION

B = μ₀I/(2πr). If I → 2I and r → r/2, then B_new = μ₀(2I)/(2π(r/2)) = 4·μ₀I/(2πr) = 4B_initial

Test
Q.195 Medium Magnetism
The permeability of free space μ₀ has the value:
A 8.85 × 10⁻¹² F/m
B 4π × 10⁻⁷ T·m/A
C 3 × 10⁸ m/s
D 6.63 × 10⁻³⁴ J·s
Correct Answer:  B. 4π × 10⁻⁷ T·m/A
EXPLANATION

The permeability of free space μ₀ = 4π × 10⁻⁷ T·m/A. Option A is permittivity ε₀, option C is speed of light, and option D is Planck's constant.

Test
Q.196 Medium Magnetism
A charged particle with charge q and mass m is moving with speed v in a circular path of radius r in a magnetic field. The magnetic field strength is:
A B = mv/(qr)
B B = qr/(mv)
C B = qv/(mr)
D B = mr/(qv)
Correct Answer:  A. B = mv/(qr)
EXPLANATION

From qvB = mv²/r (centripetal force equals magnetic force), we get B = mv/(qr). This is the relationship between field strength, particle properties, and circular path radius.

Test
Q.197 Medium Magnetism
The magnetic susceptibility of a paramagnetic material is:
A Negative and large in magnitude
B Positive and small
C Zero
D Negative and small
Correct Answer:  B. Positive and small
EXPLANATION

Paramagnetic materials have positive but small magnetic susceptibility (χ > 0, typically 10⁻⁵ to 10⁻³). Diamagnetic materials have small negative susceptibility, and ferromagnetic materials have large positive susceptibility.

Test
Q.198 Medium Magnetism
A conducting rod of length L moves with velocity v perpendicular to its length in a magnetic field B. The motional EMF induced is maximum when:
A v is parallel to B
B v is perpendicular to B
C v is at 45° to B
D B is zero
Correct Answer:  B. v is perpendicular to B
EXPLANATION

Motional EMF = B·L·v·sinθ, where θ is the angle between v and B. EMF is maximum when sinθ = 1, i.e., when v is perpendicular to B.

Test
Q.199 Medium Magnetism
Two magnets are placed with their north poles facing each other. The force between them varies with distance r as:
A F ∝ 1/r
B F ∝ 1/r²
C F ∝ 1/r³
D F ∝ 1/r⁴
Correct Answer:  D. F ∝ 1/r⁴
EXPLANATION

Two magnetic dipoles interact with force F ∝ 1/r⁴ when aligned along the same axis. This is because the magnetic field of a dipole varies as 1/r³, and force on a dipole is proportional to the field gradient.

Test
Q.200 Medium Magnetism
A rectangular conducting loop ABCD with sides a and b is rotated with angular velocity ω in a uniform magnetic field B perpendicular to the plane of rotation. The induced EMF is:
A Bab·ω
B Bab·ω·sinωt
C Bab·ω·cosωt
D Bab·ω/(sinωt)
Correct Answer:  B. Bab·ω·sinωt
EXPLANATION

When the loop rotates, the magnetic flux through it varies as Φ = BA·cosωt. The induced EMF = -dΦ/dt = BA·ω·sinωt = Bab·ω·sinωt

Test
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